In summary, two metal spheres of equal radius R are initially placed at a large distance from each other, with one sphere having a charge q and the other being uncharged. The spheres are then brought into contact and then separated again. The task is to find the external work done during this process. Using the equation for electrostatic potential energy of a sphere, the external work can be calculated as either positive or negative depending on the perspective. However, since the energy of the system decreases during the process, it can be concluded that the external work done is negative.
  • #1
crick
43
4

Homework Statement


Two metal spheres of equal radius ##R## are placed at big distance one from the other. Sphere 1 has total charge ##q## and sphere 2 has no charge. The two speheres are moved one towards the other until they touch, then they are moved again far away one from the other. What is the external work ##W_{EXTERNAL \,\, FORCES}## done during the process?

Homework Equations


Electrostatic potential energy of a sphere is
##U_{EL}=\frac{1}{2} \frac{q^2}{4 \pi \epsilon_0 R}##

The Attempt at a Solution


I was convinced that, since the only forces are the electric forces and the external forces, $$W_{ELECTRIC \,\, FORCES}= - \Delta U_{EL}= U_{EL \,\, initial}-U_{EL \,\, final}=\frac{1}{2} \frac{q^2}{4 \pi \epsilon_0 R}-2 \cdot \frac{1}{2} \frac{(q/2)^2}{4 \pi \epsilon_0 R}$$
So the work of external forces, which is what is asked, is
$$W_{EXTERNAL\,\, FORCES}= \Delta U_{EL}=U_{EL \,\, final}- U_{EL \,\, initial}=2 \cdot \frac{1}{2} \frac{(q/2)^2}{4 \pi \epsilon_0 R}-\frac{1}{2} \frac{q^2}{4 \pi \epsilon_0 R}$$

But solutions says exactly the opposite, that is

$$W_{EXTERNAL \,\, FORCES}= - \Delta U_{EL}= U_{EL \,\, initial}-U_{EL \,\, final}=\frac{1}{2} \frac{q^2}{4 \pi \epsilon_0 R}-2 \cdot \frac{1}{2} \frac{(q/2)^2}{4 \pi \epsilon_0 R}$$

How can this be true? Am I missing something on the sign or is the solution wrong?
 
Physics news on Phys.org
  • #2
When the spheres make contact, work is lost, so you cannot get the work done by external forces from the net change in energy of the spheres.
 
  • Like
Likes crick
  • #3
Thanks for the reply! I see that there are some problems in that contact with the conservation of energy, but I'm a bit confused. Which work is lost when there is contact? The external forces or the electrostatic forces one?

So what part of potential energy should I consider?
After the contact the potential energy of the spheres does not change anymore, so I'm not even sure that any work is done..
 
  • #4
crick said:
Thanks for the reply! I see that there are some problems in that contact with the conservation of energy, but I'm a bit confused. Which work is lost when there is contact? The external forces or the electrostatic forces one?

So what part of potential energy should I consider?
After the contact the potential energy of the spheres does not change anymore, so I'm not even sure that any work is done..
I've thought about this some more, and now think I was not helpful.
In a practical setting there would be some work lost, but in a theoretical situation, with no leakage right up until the instant of contact, there might not be. When in contact, obviously there will be Q/2 on each, in some symmetrical distribution. Perhaps the distribution immediately prior to contact is the same, with the addition of Q/2 on the charged sphere as almost a point charge at the place contact is about to be made, and a corresponding -Q/2 on the other. This would give a constant potential on each sphere, and in the limit the two would be the same.
When contact occurs, the two almost point charges neutralise, but since they had hardly any separation scarcely any work is lost.

So, back to your question. Clearly, the energy of the system has reduced, so the work done by the external forces must be negative. This agrees with your solution. But I note the wording:
crick said:
What is the external work ... done during the process?
I can read that either way: the work done by external forces or the work done by the system on external entities. Is that the exact wording?
 

1. What is the work of external forces on an object?

The work of external forces on an object is the force exerted on the object multiplied by the distance the object moves in the direction of the force. This work can either increase or decrease the object's kinetic energy, depending on the direction of the force.

2. How is the work of external forces related to electrostatic potential energy?

Electrostatic potential energy is the potential energy stored in a system due to the interactions between charged particles. The work of external forces, specifically those caused by electric fields, can change the electrostatic potential energy of a system by either increasing or decreasing the distance between charged particles.

3. Can the work of external forces on an object be negative?

Yes, the work of external forces can be negative. This occurs when the force is acting in the opposite direction of the object's motion, causing a decrease in kinetic energy. An example of this is when an object is moving against the force of friction, which causes the object's motion to slow down.

4. How is electrostatic potential energy different from other forms of potential energy?

Electrostatic potential energy is different from other forms of potential energy because it is due to the interactions between charged particles. Other forms of potential energy, such as gravitational potential energy, are due to the position of an object in a force field.

5. Can the work of external forces on an object change the object's electrostatic potential energy?

Yes, the work of external forces can change an object's electrostatic potential energy. As mentioned earlier, the work of external forces, specifically those caused by electric fields, can change the distance between charged particles and therefore change the electrostatic potential energy of a system.

Similar threads

  • Introductory Physics Homework Help
Replies
23
Views
354
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
699
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
361
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
413
  • Introductory Physics Homework Help
Replies
22
Views
417
  • Introductory Physics Homework Help
Replies
12
Views
748
Back
Top